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Question Number 163496 by Ahmed777hamouda last updated on 07/Jan/22
∫0π2256cos5(x2)sin11(x2)dx
Answered by Ar Brandon last updated on 07/Jan/22
I=∫0π2256cos5(x2)sin11(x2)dx=8∫0π2sin5x(1−cosx2)3dx=∫0π2sin5x(1−3cosx+3cos2x−cos3x)dx=12β(3,12)−32β(3,1)+32β(3,32)−12β(3,2)=25⋅23⋅21−12+32⋅27⋅25⋅23⋅21−14×3×2×1=815−12+835−124=37168
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